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3 years ago

Given the graph of a function below, which of the following points would be on the graph of the inverse of the function?

Mathematics

1 answer:

pshichka [43]3 years ago

4 0

Its the second one. The inverse is the opposite of the original

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A circular picture is 8 inches in diameter.

velikii [3]

The formula of an area of a circle with radius r:

$A=\pi r^2$

Part A.

We have diameter of the circle d = 8 in. The diameter of a circle is equal two times length of a radius.

Therefore:

$r=\dfrac{d}{2}\to r=\dfrac{8}{2}=4\ in$

Calculate the area:

$A=\pi\cdot4^2=16\pi\ in^2$

Answer: C. 16π square inches.

Part 2.

Look at the picture.

Total area:

$A=\pi\cdot6^2=36\pi\ in^2$

Answer: C. 36π square inches

7 0

3 years ago

The Junior Beta club needs to raise $1513.75 to go to a national convention. If they decided to sell candy bars at $1.25 each, h

rusak2 [61]

Answer:

1,211 candy bars.

Step-by-step explanation:

You divide the goal by how much the bars are selling for, or 1513.75/1.25. The answer to this is 1,211 candy bars.

4 0

3 years ago

Read 2 more answers

Integrate Sec (4x - 1) Tan (4x - 1) Dx

Delicious77 [7]

$\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill$

$\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C$

5 0

3 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

The area of rectangle is equal to

$A=14*5=70\ in^{2}$

The area of the small semicircle is equal to

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

The area of the larger semicircle is equal to

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

A growing farming conglomerate increases its water usage at a rate of 5% every year. If it used 48,000 megaliters of water this

romanna [79]

Answer:

Future requirement of water = 115,517 mega-liters

Step-by-step explanation:

Given:

Current requirement of water (C) = 48,000 mega-liters

Increasing rate per year (g) = 5% = 0.05

Number of year (n) = 18 years

Find:

Future requirement of water = ?

Computation:

$Future\ requirement\ of\ water = C(1+g)^n\\\\Future\ requirement\ of\ water = 48,000(1+0.05)^{18}\\\\ Future\ requirement\ of\ water = 48,000(1.05)^{18}\\\\ Future\ requirement\ of\ water = 48,000(2.4)\\\\Future\ requirement\ of\ water = 115,517 (Approx)$

Future requirement of water = 115,517 (Approx) mega-liters

7 0

3 years ago